Question

If a scuba diver descends too quickly into the sea, the internal pressure on each eardrum remains at atmospheric pressure, while the external pressure increases due to the increased water depth. At sufficient depths, the difference between the external and internal pressures (the internal pressure can be assumed to be atmospheric pressure) can rupture an eardrum. eardrum can rupture to different degrees when the pressure difference is as little as 20kPa-35 kPa, depending on the person. Suppose that the danger level for eardrum rupture for a particular scuba diver is a pressure difference of 22,964 N/m². What is the depth at which this pressure difference could occur? The density of seawater is 1025 kg/m³.

Answer 1

The danger level for eardrum rupture for the given scuba diver, which corresponds to a pressure difference of 22,964 N/m², occurs at a depth of approximately 12.43 meters in seawater with a density of 1025 kg/m³.

Step-by-step explanation:

To calculate the depth at which a pressure difference of 22,964 N/m² could rupture a scuba diver's eardrum, we can use the formula for pressure due to a fluid column, which is P = ρgh, where P is the pressure, ρ (rho) is the density of the fluid, g is the acceleration due to gravity, and h is the height or depth of the fluid column.

First, we add atmospheric pressure to the pressure difference the eardrum can tolerate since the total pressure exerted on the diver is the sum of atmospheric pressure and the pressure from the water column. Atmospheric pressure at sea level is approximately 101,325 N/m². Therefore, the total pressure at the depth where the eardrum will rupture is 101,325 N/m² + 22,964 N/m² = 124,289 N/m².

To find the required depth (h), we rearrange the formula to h = P / (ρg). Using the given density of seawater ρ = 1025 kg/m³ and assuming g = 9.81 m/s², we get:

h = 124,289 N/m² / (1025 kg/m³ · 9.81 m/s²) = 12.43 m

This means that the danger level for eardrum rupture for this particular scuba diver is at a depth of about 12.43 meters.